By the induction assumption,
$73 | 8^{n+2}+9^{2n+1}$,
so $73$ divides this sum,
and therefore divides
$8^{n+3}+9^{2n+3}$.

I worked this out as I went along,
and the only thing that might be considered
"clever"
is noticing that
$80 = 73+7$.
I guess that
I "had" to do something at that point,
because otherwise I would have been stuck.
This also forced the induction hypothesis
to come into play here.
I find this interesting because it
is used twice,
once in computing the difference,
and once in showing that
the difference is divisible by 73.

It might be interesting to try
to generalize this,
because this kind of thing
is rarely a coincidence.

$$8\cdot 8^{k+2}+81\cdot 9^{2k+1}=8\cdot(8^{k+2}+9^{2k+1})+?$$ is one example.

Note also that any sequence of the form $a_n=r\cdot 8^n+s\cdot81^n$ satisfies a recurrence relation i.e. $$a_{n+2}-(8+81)a_{n+1}+8\cdot 81 a_n=0$$ where you can see how the coefficients are made up. Then you need two cases to get started, but it is obvious that divisibility persists.

Hint: Try adding and subtracting $8\cdot 9^{2k+1}$ (or $81\cdot 8^{k+2}$) and see what happens. The reason for doing this is that, as written, we can't immediately invoke the inductive hypothesis but if we introduce either of these terms, we would get exactly what we want.